Page:LorentzContraction1921.djvu/3

(system $S$), and $l_{0}$ their distance in the natural state (system $S_{0}$), these distances being both determined in the manner specified in what precedes. Then

$\xi ^{\xi }=\left(l-l_{0}\right)/l_{0}\,$

Again, if $P''$ is the particle $\xi ,\eta +d\eta ,\zeta$, and if the angle $P'PP''$, calculated as stated before, has in the two cases the values $\delta$ and $\delta _{0}\left(={\tfrac {1}{2}}\pi \right)$, we shall have

$\xi ^{\eta }=\delta _{0}-\delta$

The six deformations $\xi _{\xi }\dots$ will be considered as infinitely small. In the problem we have in view, they are of the order of magnitude $v^{2}/c^{2}$, so that our final result will be correct to that order.

If we put

$U=A\left(\xi _{\xi }^{2}+\eta _{\eta }^{2}+\zeta _{\zeta }^{2}\right)+{\frac {1}{2}}B\left(\xi _{\xi }+\eta _{\eta }+\zeta _{\zeta }\right)^{2}+{\frac {1}{2}}A\left(\xi _{\eta }^{2}+\eta _{\zeta }^{2}+\zeta _{\xi }^{2}\right)$

(6)

the well-known expression for the potential energy of an isotropic elastic body, $U$ will be invariant for any change of co-ordinates.

As to the kinetic energy $T$, it is to be replaced by an expression containing $\rho {\tfrac {ds}{dt}}$. Finally, we must write, instead of (3),

$\delta \int _{t_{1}}^{t_{2}}\int \left(-c\rho -{\frac {1}{c}}U\right){\frac {ds}{dt}}dt\ d\xi \ d\eta \ d\zeta =0$

We have still to add the formula; that are found by working out the above definitions of $\xi _{\xi },\xi _{\eta }$, etc. viz.:—

$\xi _{\xi }=-{\frac {1}{2}}\sum (\alpha \beta )g_{\alpha \beta }{\frac {\partial x_{\alpha }}{\partial \xi }}{\frac {\partial x_{\beta }}{\partial \xi }}+{\frac {\left[\sum (\alpha \beta )g_{\alpha \beta }v_{\beta }{\frac {\partial x_{\alpha }}{\partial \xi }}\right]^{2}}{2\sum (ab)g_{ab}v_{a}v_{b}}}-{\frac {1}{2}}$ $\xi _{\eta }=-\sum (\alpha \beta )g_{\alpha \beta }{\frac {\partial x_{\alpha }}{\partial \xi }}{\frac {\partial x_{\beta }}{\partial \eta }}+{\frac {\sum (\alpha \beta )g_{\alpha \beta }v_{\beta }{\frac {\partial x_{\alpha }}{\partial \xi }}\cdot \sum (\alpha \beta )g_{\alpha \beta }v_{\beta }{\frac {\partial x_{\alpha }}{\partial \eta }}}{\sum (ab)g_{ab}v_{a}v_{b}}}$

($v_{1},v_{2},v_{3}$ are the components of the velocity, and $v_{4}=1$).

In our problem the body is supposed to move in a normal system of co-ordinates. By this our formula; simplify to^[1]

$\delta \int _{t_{1}}^{t_{2}}\int \left(-c^{2}\rho -U\right)\left(1-{\frac {v^{2}}{c^{2}}}\right)^{\frac {1}{2}}dt\ d\xi \ d\eta \ d\zeta =0,$ $\xi _{\xi }={\frac {1}{2}}\sum (\alpha )\left({\frac {\partial x_{\alpha }}{\partial \xi }}\right)^{2}+{\frac {\left[\sum (\alpha )v_{a}{\frac {\partial x_{\alpha }}{\partial \xi }}\right]^{2}}{2\left(c^{2}-v^{2}\right)}}-{\frac {1}{2}},$ $\xi _{\eta }=\sum (\alpha ){\frac {\partial x_{\alpha }}{\partial \xi }}{\frac {\partial x_{\alpha }}{\partial \eta }}+{\frac {\sum (\alpha )v_{\alpha }{\frac {\partial x_{\alpha }}{\partial \xi }}\cdot \sum (\alpha )v_{a}{\frac {\partial x_{\alpha }}{\partial \eta }}}{c^{2}-v^{2}}},$

(7)

When applied to a revolving body, these equations will enable us to determine the deformation that is produced, wholly independently of the theory of relativity, by centrifugal force, a deformation that will in reality far surpass the changes we want to consider. To get free from it we can consider the ideal case of a "rigid" body — i.e. a body for which the moduli of elasticity $A$ and $B$ in (6) are infinitely great. The centrifugal force will then have no effect on the dimensions, but the changes required by the theory of relativity will subsist. The assumption has also the advantage of simplifying the calculations ; indeed, since $U$ becomes infinitely great, the term $-c^{2}\rho$ in (7) may be omitted.

I have worked out the case of a thin circular disc rotating with constant speed about an axis passing through its centre, at right angles to its plane. The result is that, if $v$ is the velocity at the rim, the radius will be shortened in the ratio of 1 to $1-{\tfrac {1}{8}}{\tfrac {v^{2}}{c^{2}}}$. The circumference changing to the same extent, its decrease is seen to be exactly one-fourth of that of a rod moving with the same velocity in the direction of its length. That there would be a smaller contraction was to be expected; indeed, the case can be compared to what takes place when a hot metal band is fitted tightly around a wheel and then left to cool.

At first sight our problem seems to lead to a paradox. Let there be two equal discs $A$ and $B$, mounted on the same axis, $A$ revolving and $B$ at rest. Then $A$ will be smaller than $B$, and it must certainly appear so (the discs being supposed to be quite near each other) to any observer, whatever be the system of co-ordinates he chooses to use. However, we can introduce a system of co-ordinates S' revolving with the disc $A$; with respect to these it will be $B$ that rotates, and so one might think that now this latter disc would be the smaller of the two. The conclusion would be wrong because the system $S'$ would not be a normal one. If we leave $S$ for it, we must at the same time change the potentials $g_{ab}$, and if this is done the fundamental equation will certainly again lead to the result that $A$ is smaller than $B$.

no. 2677, vol. 106]

1. If in (7) we replace $\left(1-v^{2}/c^{2}\right)^{\frac {1}{2}}$ by $1-v^{2}/2c^{2}$, omitting the constant term $-c^{2}\rho$ and neglecting $U\cdot v^{2}/2c^{2}$, we are led back to the ordinary formula (3).